A function `y = f(x)` satisfies the condition `f'(x) sin x + f(x) cos x=1 ,f(x)` being bounded when `x->0`. If `I= int_0^(pi/2) f(x) dx` then (A) `pi/2ltIltpi^2/4` (B) `pi/4ltIltpi^2/2` (C) `1ltIltpi/2` (D)`0ltIlt1`

Options

A. `(pi)/(2)ltIlt(pi^(2))/(4)`
B. `(pi)/(4)ltIlt(pi^(2))/(2)`
C. `1ltIlt(pi)/(2)`
D. `0ltIlt1`

Answer

To solve the problem, we start with the given differential equation:

\[ f'(x) \sin x + f(x) \cos x = 1 \]

### Step 1: Rewrite the equation
We can rewrite the left side of the equation as the derivative of a product:

\[ \frac{d}{dx} (f(x) \sin x) = 1 \]

### Step 2: Integrate both sides
Now we integrate both sides with respect to \(x\):

\[ \int \frac{d}{dx} (f(x) \sin x) \, dx = \int 1 \, dx \]

This gives us:

\[ f(x) \sin x = x + C \]

where \(C\) is the constant of integration.

### Step 3: Solve for \(f(x)\)
Now, we can solve for \(f(x)\):

\[ f(x) = \frac{x + C}{\sin x} \]

### Step 4: Analyze the bounded condition
We know that \(f(x)\) is bounded as \(x \to 0\). We need to analyze the behavior of \(f(x)\) as \(x\) approaches 0:

Using the small-angle approximation, \(\sin x \approx x\) as \(x \to 0\):

\[ f(x) \approx \frac{x + C}{x} = 1 + \frac{C}{x} \]

For \(f(x)\) to be bounded as \(x \to 0\), \(C\) must be 0. Thus, we have:

\[ f(x) = \frac{x}{\sin x} \]

### Step 5: Evaluate the integral \(I\)
Now we need to evaluate the integral:

\[ I = \int_0^{\frac{\pi}{2}} f(x) \, dx = \int_0^{\frac{\pi}{2}} \frac{x}{\sin x} \, dx \]

### Step 6: Analyze the bounds of \(I\)
We know that for \(0 < x < \frac{\pi}{2}\), \(\sin x < x\), which implies:

\[ \frac{x}{\sin x} > 1 \]

Thus:

\[ I > \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \]

Now, we also know that \(\frac{x}{\sin x} < \frac{\pi/2}{\sin(\pi/2)} = \frac{\pi}{2}\), which implies:

\[ I < \int_0^{\frac{\pi}{2}} \frac{\pi/2}{\sin x} \, dx \]

However, we can also use the fact that \(\frac{x}{\sin x} \to 1\) as \(x \to 0\) and is continuous and increasing in the interval \((0, \frac{\pi}{2})\).

### Conclusion
From the analysis, we conclude:

\[ \frac{\pi}{2} < I < \frac{\pi^2}{4} \]

Thus, the correct option is:

**(A) \(\frac{\pi}{2} < I < \frac{\pi^2}{4}\)**